Download CYL100 2013–14 I semester Homework 2 Solutions 1. Consider a

CYL100 2013–14 I semester Homework 2 Solutions
1. Consider a well-insulated piston-cylinder assembly. On the 0.05 m2 piston rests two 5000-kg blocks. The initial
temperature is 500 K. The ambient pressure is 5 bar. One mole of an ideal gas is contained in the cylinder. The
gas is compressed in a process in which another 5000-kg block is added. The heat capacity at constant volume can
be taken to have a constant value of (5/2)R.
(a) What are the initial and final pressures of the gas in the system?
P1 = Psurr +
2(5000kg)(9.81m/s2 )
= 24.6 × 105 P a
0.05m2
P1 = Psurr +
2(5000kg)(9.81m/s2 )
= 24.6 × 105 P a
0.05m2
(b) Do you expect the temperature to rise or fall? Explain.
Temperature should rise - convention we have followed is that work done on the system is positive. System is
insulated so energy conservation would require the temperature to rise.
(c) What is the final temperature?
∆U = w; nCv (T2 − T1 ) = −P2 (V2 − V1 )
n(5/2R)(T2 − T1 ) = −P2 (
nRT2
− V1 ); T2 = 557K]
P2
(d) Calculate ∆Ssys and ∆Ssurr .
∆Ssurr = 0; ∆Ssys = Cv ln
V2
T2
R ln
= 0.354J/molK
T1
V1
(e) Does this process violate the second law of thermodynamics? Explain.
No. ∆Stotal = ∆Ssys > 0.
2. For the process given in the problem above, calculate ∆Ssys using each of the following paths:
Without calculation we know that we should get the same result as above for these three paths between the same
initial and final state.
(a) a reversible, adiabatic compression, followed by a reversible, isothermal expansion
(Pi , Ti ) = (24.6bar, 500K) → (P2 , Tf ) = (P2 , 557K) → (Pf , Tf ) = (34.4bar, 557K)
Entropy in the first step is zero - adiabatic reversible compression.
P2 =
T1
T2
γ
(1−γ)
Pi = 35.9bar
∆S = −R ln P2 /P1
(b) a reversible, isobaric heating followed by a reversible, isothermal compression
(Pi , Ti ) = (24.6bar, 500K) → (24.6bar, 557K) → (Pf , Tf ) = (34.4bar, 557K)
Entropy change during a reversible, isobaric heating is ∆S = CP ln Tf /Ti while it is ∆S = −R ln Pf /Pi =
R ln Vi /Vf during a reversible isothermal compression.
(c) a reversible, isochoric heating followed by a reversible, isothermal compression
(Pi , Ti ) = (24.6bar, 500K) → (P2 , Tf ) = (P2 , 557K) → (Pf , Tf ) = (34.4bar, 557K)
P1
P2
=
T1
T2
Entropy change during a reversible, isochoric heating is ∆S = Cv ln Tf /Ti while it is ∆S = −R ln Pf /P2 =
R ln Tf /Ti − R ln Pf /Pi during a reversible isothermal compression.
3. A salesperson claims that a mysterious black box, with no moving parts, can take an inlet stream of ideal gas at 2
kg/s and 4 bar and 50 ℃ and cool 0.5 kg/s of it to −10 ℃ and 1 bar while the rest of the gas is raised to 70 ℃ at
1 bar. Is this possible? Explain.
(i) Mass is conserved
(ii) Energy is conserved: 0.5kg/sCP (−10 − 50)K = 1.5kg/sCP (70 − 50)K
(iii) Does the second law hold? For an adiabatic process,
dS
= 0.5kg/s(S2 − S1 ) + 1.5kg/s(S3 − S1 )
dt univ
S2 − S1 = Cp ln T2 /T1 − R ln P2 /P1
S3 − S1 = Cp ln T3 /T1 − R ln P3 /P1
dS
= 0.5kg/sCP (ln 263 + 3 ln 343 − 4 ln 323) + 2R ln 4
dt univ
Second law holds as long as CP ≤ 218R, which is true for all imaginable gases.
4. Calculate the maximum work and the maximum non-expansion work that can be obtained from the freezing of
supercooled water at −5 ℃ and 1.0 atm. The densities of water and ice are 0.999 and 0.917 g cm−3 , respectively
at −5 ℃.
The simplest procedure is to calculate the change in G and A by using the relationship ∆A = ∆U − T ∆S amd
∆G = ∆H − T ∆S. From thermodynamic tables we know that ∆H = −6008 J/mol for freezing at 0 ℃, CP for
liquid water is 75 J/(mol K), and CP for ice is 36.4 J/(mol K). The computation of ∆H and ∆S was discussed in
class. As a zeroth approximation, we could take that ∆U = ∆H because condensed phases are involved. We could
do better by using the densities to find the P dV contribution to ∆H.
∆H(268K) = ∆H(273K) + (CP (solid) − CP (liquid))(−5K)
∆U (268K) = ∆H(268K) − 1atm(Vm (solid) − Vm (liquid))
∆S(268K)
5. One mole of He is heated from 200 ℃ to 400 ℃ at a constant pressure of 1 atm. Given that the absolute entropy
of He at 200 ℃ is 810 JK−1 mol−1 , and assuming He is a perfect gas, comment on the spontaneity of the process.
∆S = S0 + CP ln 673/473
The spontaneity of the process will be determined by how the heating is done. If the heating is achieved by putting
the gas into a heat bath at 400 ℃, then ∆Stot = ∆Ssys + ∆Ssurr > 0.
∆Ssurr = −∆Hsys /673;
6. Derive the relations: (i) Cp − Cv = T
∂V
∂T
∂P
∂T V
P
∂H
∂T
∆Hsys = CP (200K)
;
−
P
∂U
∂T
V
Use the definition H = U + P V
∂H
∂U
∂U
∂V
∂U
−
=
+P
−
∂T P
∂T V
∂T P
∂T P
∂T V
To obtain an expression for ∂U
∂T P we proceed as follows
∂U
∂U
dU =
dT +
dV
∂T V
∂V T
Dividing by dT and applying constant pressure gives
∂U
∂U
∂U
∂V
=
+
.
∂T P
∂T V
∂V T ∂T P
Thus we have
CP − CV =
∂U
∂V
T
∂V
∂T
+P
P
Using the result that was proved in class
∂U
∂V
=T
T
∂P
∂T
−P
V
∂V
∂T
P
yields what we set out to prove.
2
(ii) Cp − Cv = α βT V
After all the effort of the previous part, this is straightforward. Use the definition of α = V1 ∂V
∂T P and β =
∂V
∂P
∂T
− V1 ∂V
∂P T and the cyclic relation ∂P T ∂T V ∂V P = −1 in the result that was derived in part (i).
(iii) µJT = −(V/Cp )(βCv µJ − βP + 1)
∂T
and µJ =
The toughest part is to look up the definitions of µJT and µJ . µJT = ∂P
H
∂H
(iv) ∂V S = γ/β
This is a slightly tricky one. We first use the relation
∂H
∂H
∂P
=
∂V S
∂P S ∂V S
From the relation dH = T dS + V dP we know that
∂H
∂P
∂T
∂V
U
=V
S
Use the cyclic relation to write
Now rewrite
and
∂P
∂V
=−
S
∂V
∂S
∂S
∂P
=
P
Use the fact that
and
=
V
∂T
∂S
1
∂V
∂S P
∂V
∂T
∂S
∂T
P
=
P
V
∂S
∂P V
∂T
∂S
∂T
∂P
P
V
T
CP
∂S
CV
=
∂T V
T
∂V
∂T
∂V
=−
∂T P ∂P P
∂P T
to obtain the stated relation.
CV β
(v) ∂V
∂T P = T α .
This looks suspicious. My guess is that it is wrong. The dimensions on both sides seem to match. The left hand
2
side is V α implying that CV = T Vβα . This cannot be correct because in part (ii) we proved that this quantity is
CP − CV . Oops! Sorry!
7. (a) Calculate the change in the chemical potential of a perfect gas when its pressure is increased isothermally from
1.8 atm to 29.5 atm at 40 ℃.
∆µ = RT ln
P2
P1
(b) The molar Gibbs energy of a certain gas is given by
1
1
Gm = RT ln p + A + Bp + Cp2 + Dp3
2
3
where A, B, C, D are constants. Obtain the equation of state of the gas.
Vm =
∂G
∂P
=
T
RT
+ B + CP + DP 2
P
8. At 1 atm the Srh → Smon transition takes place at 95.5 ℃, and the melting point of Smon is 119.3 ℃. The latent
heat of the rhombic to monoclinic transition is 1.16104 J kg−1 and the latent heat of fusion of Smon is 5.53 × 104
J kg−1 . The densities of rhombic, monoclinic, and liquid sulphur are 2.07 × 103 , 1.96 × 103 , 1.90 × 103 kg m−3 .
Estimate the rhombic, monoclinic, and liquid triple point.
Use the Clapeyron equation to determine the slopes of the two lines. We also know that the rhombic to monoclinic
line passes through (95.5 ℃, 1 atm). Similarly for the monoclinic melting. The intersection of these two lines gives
the triple point.
9. The vapor pressure of zinc varies with temperature as
log p(mmHg) = −
6850
− 0.755 log T + 11.24
T
and that of liquid zinc as
log p(mmHg) = −
6620
− 1.255 log T + 12.34.
T
Calculate (a) the boiling point of zinc
At the boiling point, the vapor pressure of the liquid becomes the atmospheric pressure, which is 760 mm of Hg.
log 760 = −
6620
− 1.255 log T + 12.34
T
We solve this by the method of successive approximation. I drop the log T term to begin with and solve for T . This
gives me an initial guess for the root to be T = 699.85. I use this guess root in the log T and solve the equation
log 760 = −
6620
− 1.255 log 699.85 + 12.34
T
for T , which yields T = 1124.1. I repeat this process a few times and find that T = 1181.
There are other methods to solve this - graphical and using a computer algebra system are two possibilities.
(b) the triple point
At the triple point, the solid-vapor and liquid-vapor coexistence curves intersect. This means
−
6850
6620
− 1.255 log T + 12.34 = −
− 0.755 log T + 11.24
T
T
230
− 0.5 log T + 1.10 = 0
T
Solving this equation we find that the temperature at the triple point is close to 700. One uses this T in one of the
equations to find the P .
(c) the heat of evaporation at the boiling point
6620
2.303 log p(mmHg) = ln P = 2.303 −
− 1.255 log T + 12.34.
T
6620 − 1.255T
6620 1.255
−
= 2.303R
dT
d ln P = 2.303RdT
RT 2
RT
RT 2
Comparing this with the Clausius-Clapeyron equation, d ln P = ∆H ◦ /RT 2 dT , shows that ∆H ◦ = 2.303R(6620 −
1.255T ). Substitute the boiling point T = 1181 in ∆H ◦ = 2.303R(6620 − 1.255T ) and obtain ∆H ◦ .
(d) the heat of fusion
The heat of fusion, which is the enthalpy of melting, may be obtained from subtracting the heat of evaporation
from the heat of sublimation. The heat of sublimation is obtained by following the same procedure as in part (c)
for the sublimation curve.
(e) the difference in the Cp s of solid and liquid zinc.
If ∆H ◦ has the form A + BT , then it implies that B = CP . From the temperature dependent part of ∆H ◦ for
evaporation and sublimation, we can obtain the difference in the CP ’s.
10. The Planck function defined as
H
+S
T
is a thermodynamic potential similar to the Gibbs energy. Obtain a “T dS” type equation for dY and from it a
Maxwell relation. Also, what is the thermodynamic criterion for spontaneity in terms of Y .
Y =−
dH
H
1
H
dT −
+ dS = 2 dT − (dH − T dS)
T2
T
T
T
We know that dH = T dS − V dP so that
H
V
dY = 2 dT − dP
T
T
Comparing this with
∂Y
∂Y
dY =
dT +
dP
∂T P
∂P T
dY =
suggests that
∂Y
∂T
=
P
H
T2
and
∂Y
∂P
=−
T
V
.
T
Because Y is a state function, we know
∂2Y
=
∂T ∂P
∂ V
∂ H
=
∂P T 2
∂T T
1 ∂H
1 ∂V
V
−
=
T 2 ∂P T
T2
T ∂T P
∂H
∂V
=V −T
,
∂P T
∂T P
∂2Y
∂P ∂T
which is a result we derived earlier by another approach.
We see that Y = − G
T , which suggests that (dY )T,P ≥ 0 would be the condition for spontaneity and equilibrium.
11. It is often claimed that one can skate on ice because the pressure of the skate causes the ice to melt, thus dramatically
reducing the friction between skate and ice. While this makes a good story, is this quite correct? Consider a skater
with a mass of 75 kg on a skate that is 3 mm wide and 20 cm long. The densities of water and ice are 0.999 and
0.917 g cm−3 respectively, and the heat of fusion is 6.004 kJ mol−1 . Since common experience is that ice skating
is possible even when the ambient temperature is well below the normal freezing point, does the pressure induced
lowering of the melting point explain clearly this observation?
Straightforward implementation of the Clapeyron equation (watch out for the units and remember that we need
the molar volume) using the data given
∆H
6004
dP
=
=
1
dT
T ∆V
2730.018 999
−
1
917
= 1.3649 × 107 P a
shows that it requires more than 130 atm to decrease the melting point by 1 K at 273K. The pressure exerted by
the skater is a little over an atmosphere so that the change in melting point as a result of the pressure on the skates
is at most a few hundredths of a degree. The pressure induced lowering of the melting point clearly does not explain
this every day observation.